Gen. Math. Notes, Vol. 23, No. 2, August 2014, pp.16-27
ISSN 2219-7184; Copyright c ICSRS Publication, 2014
www.i-csrs.org
Available free online at http://www.geman.in
Adding a General Union of a Prescribed Number
of Curves with High Sum of their Degrees Improve
the Hilbert Function of any Scheme
Edoardo Ballico
Department of Mathematics, University of Trento
38123 Povo (TN), Italy
E-mail: ballico@science.unitn.it
(Received: 6-4-14 / Accepted: 15-5-14)
Abstract
Let Z ⊂ P , r ≥ 4, be a closed subscheme with dim(Z) ≤ r −4. Fix integers
c > 0 and gi ≥ 0, i = 1, . . . , c. We prove that the general union of Z and c
smooth curves Yi ⊂ Pr with genus gi and deg(Yi ) ≥ r + gr as maximal rank
(i.e. the expected postulation) if deg(Y1 ) + · · · + deg(Yc ) 0.
Keywords: Postulation, Hilbert function, Disjoint unions of curves
r
1
Introduction
Let W ⊂ Pr be any closed subscheme. We say that W has maximal rank if for
every integer t either h0 (IW (t)) = 0 or h1 (IW (t)) = 0, i.e. if for every integer
t the restriction map ρW,t : H 0 (OPr (t)) → H 0 (OW (t)) is either injective or
surjective, i.e. it is a linear map
with maximal rank, i.e. for all t ≥ 0 we
0
have h0 (IW (t)) = max{0, r+t
−
h
(OW (t))}. Usually the integer h0 (OW (t))
r
is known and hence maximal rank for the integer t means that the set of all
degree t hypersurfaces containing W has the “ expected ” dimension, i.e. ρW,t
has maximal rank if and only if the set of all degree t hypersurfaces containing
W has the minimal dimension which a priori may have. Set δr (0) := 1 and
δr (1) := r + 1. For all integers r ≥ 3 and g ≥ 2 set δr (g) := r + gr.
Adding a General Union of a Prescribed Number...
17
In this paper we prove the following results, easily seen to be equivalent (we
prove Theorem 2 and then prove, assuming it for fixed data r, Z, c, g1 , . . . , gc
that Theorem 1 is true for the same data, but it is very easy to see the other
implication).
Theorem 1 Fix integers r ≥ 4, c > 0 and gi ≥ 0, 1 ≤ i ≤ c. Let Z ⊂ Pr be a
closed subscheme with dim(Z) ≤ r − 4. Then there in an integer ∆ (depending
only on r, c, g1 , . . . , gc and the Hilbert polynomial of Z) with the following
property. Fix integers di ≥ δr (gi ), 1 ≤ i ≤ c, such that d1 + · · · + dc ≥ ∆. Let
Y ⊂ Pr \ Z be a general union of c smooth curves Y1 , . . . , Yc with deg(Yi ) = di
and pa (Yi ) = gi for all i. Then Z ∪ Y has maximal rank.
Theorem 2 Fix integers r ≥ 4, c > 0 and gi ≥ 0, 1 ≤ i ≤ c. Let Z ⊂ Pr be
a closed scheme with dim(Z) ≤ r − 4. Then there in an integer k0 (depending
only on r, c, g1 , . . . , gc and the Hilbert polynomial of k) with the following
property. Fix integers di ≥ δr (gi ), 1 ≤ i ≤ c. Let Y ⊂ Pr \ Z be a general
union of c smooth curves Y1 , . . . , Yc with deg(Yi ) = di and pa (Yi ) = gi for all
i. Then for all integers k ≥ k0 either h0 (IZ∪Y (k)) = 0 or h1 (IZ∪Y (k)) = 0.
The generality assumptions in Theorems 1 and 2 implies that Yi ∩ Yj for
all i 6= j and that Z ∩ Y = ∅, so that if p(t) is the Hilbert polynomial of Z,
then p(t) + c − g1 − · · · − gc + (d1 + · · · + dc )t is the Hilbert polynomial of
Z ∪ Y . As in [7] and [5] we use the so-called Horace method. As in [7] and in
several later papers we use embedded deformation of reducible curves inside a
projective space ([9],[6]).
We work over an algebraically closed field K
2
Preliminaries
For all integers r ≥ 3, g ≥ 0 and d ≥ max{2g + 1, g + r} let Z(r, d, g) denote
the set of all smooth and non-degenerate curves X ⊂ Pr with degree d, genus g
and h1 (X, OX (1)) = 0. Each set Z(r, d, g) is a non-empty and irreducible open
subset of the Hilbert scheme of Pr . This observation explains the word “ general ” in the statement of Theorems 1 and 2. Indeed, each
Qc c-ple (Y1 , . . . , Yc )
must be a general element of the irreducible algebraic set i=1 Z(r, gi , di ). Let
Z 0 (r, d, g) denote the closure of Z(r, d, g) in the Hilbert scheme of Pr .
Fix X ∈ Z(r, d, c). Let E ⊂ Pr be a smooth rational curve such that the
curve X ∪ E is nodal and connected and x := ](X ∩ E) ≤ 2. Set e := deg(E).
The curve X ∪ E has arithmetic genus g + x − 1 and degree d + e. Since
X ∪ E is nodal, its normal sheaf is locally free. We claim that X ∪ E ∈
Z 0 (r, d + e, g + e − 1); this statement is almost contained in both [6] and [9]
18
Edoardo Ballico
(both papers handle more general and difficult cases); it can be proved first
degenerating E to a connected union E 0 of e lines with E 0 ∪ X nodal and then
applying to X ∪E 0 either [9, Theorem 5.2] or [6, Theorem 4.1, Remark 4.1.2 and
Corollary 4.2]. A Mayer-Vietoris exact sequence shows that h1 (OX∪E (1)) = 0.
Hence for any Y ∈ Z(r, d + e, g + e − 1) and any integer t ≥ 0 we have
h0 (OX∪E (t)) = td + 1 − (g + x − 1) = h0 (OY (t)). Let Z ⊂ Pr be any closed
subscheme with dim(Z) ≤ r − 2 and Z ∩ (X ∪ E) = ∅. Fix t ∈ N. By the
semicontinuity theorem for cohomology ([4, Theorem III.12.8]) to prove that
either h1 (IZ∪Y (t)) = 0 or h0 (IZ∪Y (t)) = 0 for a general Y ∈ Z(r, d+e, g+x−1)
it is sufficient to prove that either h1 (IZ∪X∪E (t)) = 0 or h0 (IZ∪X∪E (t)) = 0
For each smooth curve X ⊂ Pr let NX denote its normal bundle.
Remark 1 Fin integers n ≥ 3, g ≥ 0, d ≥ g + n and x > 0. Fix any
C ∈ Z(n, d, g) and any S ⊂ C such that ](S) = x. To prove that for a general
B ⊂ Pn there is X ∈ Z(n, d, g) with B ⊂ X and that the set of all such curves
X has the expected dimension dim(Z(n, d, g))−x(n−1) it is sufficient to prove
that h1 (NC (−S)) = 0 (use the proof of [8, Theorem 1.5] or [3]). Since C is
smooth, NC is a quotient of T Pr |C and hence it is a quotient of OC (1)⊕(n+1) by
the Euler’s sequence of T Pn . Hence h1 (NC (−S)) = 0 if h1 (OC (1)(−S)) = 0.
This is true if d − x ≥ 2g − 1. In particular if g = 0, then it is sufficient
to assume d ≥ x + 1. In the case g = 1 we may use Atiyah’s classification
of vector bundles on elliptic curves and conclude, since NC (−1) is spanned
and with no trivial factor if x ≤ d. In the case g = 0 we may do a proof
which works for all d ≥ x in the following way. We first reduce to the case
x = d = n. Then we use that any two (n + 2)-ples of points of Pn in linearly
general position are projectively normal and that any n + 2 general points of
an integral and non-degenerate curve C ⊂ Pn are linearly independent.
Lemma 1 Fix a smooth and projective curve of genus g and integer d ≥ δr (g).
Let X ⊂ Pr be a general embedding of degree d of C. Then h1 (X, NX (−1)) = 0.
Proof. Since d > 2g − 2, we are looking at non-special embeddings of C
and hence the set of all such embeddings (for fixed d and r) is parametrized by
an irreducible variety. Hence the word “ general ” is allowed. If g = 0, then we
use Remark 1 (notice that in the case g = 0 we allow degenerate embeddings if
1 ≤ d < r). Now assume g ≥ 1. By the universal property of the Grassmannian
there is a bijection between morphisms f : C → Pr and pairs (E, h), where E
is a rank r vector bundle on C and h : Kr+1 → H 0 (C, E) is a linear map with
its image h(Kr+1 ) spanning E. In this bijection E ∼
= f ∗ (T Pr (−1)) and f (C)
is non-degenerate if and only h is injective. Since dim(C) = 1, a dimensional
count shows that every rank r vector bundle F on C is spanned by a linear
subspace of H 0 (C, F ) with dimension at most r + 1. Since r ≥ 3 and d ≥
r(g + 1), it is easy to find a rank r spanned vector bundle E on C for which the
Adding a General Union of a Prescribed Number...
19
map f is an embedding and h1 (C, E) = 0. For instance, take E = R1 ⊕· · ·⊕Rr
with (R1 , . . . , Rr ) general in Picd−(r−1)(g+1) (C) × Picg+1 (C)r−1 and then take a
general (r + 1)-dimensional linear subspace of H 0 (C, E) to define the injective
map h. Hence for all integers r ≥ 3 and d ≥ r(g +1) = δr (g) the general degree
d embedding X ⊂ Pr of C has the property that h1 (X, T Pr (−1)|X) = 0. Let
K ∼
= T X(−1) be the kernel of the surjection T Pr (−1) → NX (−1). Since
dim(X) = 1, we have h2 (X, K) = 0 and hence h1 (X, NX (−1)) = 0.
3
The Inductive Set-up
Set η = c − g1 − · · · − gc . Let p(t) be the Hilbert polynomial of Z. For any
hyperplane M ⊂ Pr and any closed subscheme W ⊂ Pr the residual scheme
ResM (W ) is the closed subscheme of Pr with IW : IM as its ideal sheaf. If
W = A t B with A, B closed in W and A ∩ B = ∅, then ResM (A t B) =
ResM (A) t ResM (B). If W is reduced, then ResM (W ) is the closure in Pr of
the union of all irreducible components of W not contained in M .
Let H ⊂ Pr be a hyperplane which does not contain the support of any
component of the scheme Z (i.e. of the sheaf OZ ), not even an embedded
one. By the primary decomposition theorem only finitely many subvarieties of
Zred are the support of a component of the sheaf OZ and hence, after fixing
Z, we may take as H a general hyperplane. In particular H contains no
irreducible component of Zred . Therefore Z ∩ H = ∅ if dim(Z) ≤ 0, while
dim(Z ∩ H) = dim(Z) − 1 if dim(Z) > 0. The condition that H contains
no component (not even an embedded one) of Z is equivalent to assuming
that the equation of H is a non-zero-divisor of OZ at each point of Z. Hence
ResH (Z) = Z and for each t ∈ Z the multiplication by a linear form ` with
H = {` = 0} induces an exact sequence
0 → IZ (t − 1) → IZ (t) → IZ∩H,H (t) → 0
(1)
Let p(t) be the Hilbert polynomial of Z. By (1) the Hilbert polynomial of
the scheme Z ∩ H is the first difference of p, i.e. the polynomial p1 defined
by the formula p1 (t) = p(t) − p(t − 1) for all t. For all integers r ≥ 4, all
integer-valued polynomiasl q(t) with deg(q(t)) ≤ r − 3 and all integers k > 0
define the integers ur,q(k),k and vr,q(k),k by the relations
r+k
q(k) + kur,q(k),k + vr,q(k),k =
, 0 ≤ vr,q(k),k ≤ k − 1
(2)
r
From (2) and the same equation for the integer k − 1 and another polynomial
q1 (t) we get
q(k) − q1 (k − 1) + k(ur,q(k),k − ur,q1 (k−1),k−1 )+
(3)
20
Edoardo Ballico
vr,q(k),k − vr,q(k−1),k−1)
r+k−1
=
r−1
Since deg(q(t)) ≤ r − 3, for k 0 (depending only on q(t) and r) we have
r−1
kr−2
.
ur,q(k),k ∼ k r! and ur,q(k),k − ur,q(k−1),k−1 ∼ r·(r−2)!
We will only use the case in which q(t)−q1 (t) and q(t)−p(t) are a constant,
|q(t) − q1 (t)| ≤ c + g1 + · · · + gc and |q(t) − p(t)| ≤ c + g1 + · · · + gc . Since
0 ≤ vr,a,x ≤ x − 1, from (3) we get that if k > 3(c + g1 + · · · + gc ), then
|ur,p(k),k − ur,p(k)+x,k | ≤ 1 for all x with |x| ≤ c + g1 + · · · + gc . Hence (for fixed
r, p(t), c and g1 , . . . , gc ) there is an integer k1 such that for all integers k ≥ k1
r−1
and all such polynomials q(t), q1 (t) we have h0 (OZ (k)) = p(k), ur,q(k),k ∼ k r!
kr−2
.
and ur,q(k),k − ur,q1 (k−1),k−1 ∼ r·(r−2)!
4
Proofs of Theorems 1 and 2
If r > 4 we use induction on r to prove Theorems 1 and 2, except that in
Pr−1 we only require the case gi = 0 for all i. Recall that we assumed that
dim(Z) ≤ r − 4. Hence if r = 4, then Z ∩ H = ∅. Hence if r = 4, then we may
use Theorems 1 and 2 in H = P3 ([1]). For Theorem 2 we may take k0 = 5 if
r = 3 and Z = ∅. Hence we introduce the following notation.
Notation 1 If r = 4, then set κ := 5. If r > 4, then we assume that Theorems
1 and 2 are true in H = Pr−1 for any c, but only for genera gi = 0 and we
call κ any integer which we may take as k0 in the statement of Theorem 2 in
H for Z ∩ H, c, gi = 0 for all i.
The next lemma is Assertion Bt (and its proof as Claim 2) in part (i) of
[2, §5].
Lemma 2 There is an integer k20 ≥ κ (depending only on the integer r and
the Hilbert function p(t) of Z) such that for all integers k ≥ k20 a general union
Y ⊂ Pr of a general E ∈ Z(r, ur,p(k)+1,1 − vr,p(k)+1,1 , 0) and vr,p(k)+1,1 lines
satisfies hi (IZ∪Y (k)) = 0, i = 0, 1.
Notation 2 Fix an integer k2 ≥ max{k1 , k20 , |η| + 2}. For all i = 2, . . . , c set
ei := δr (gi ) − gi ifPgi > 0 and ei := 1 if gi = 0. Set α := ur,p(k2 +1)+c,k2 +1 −
vr,p(k2 +1)+η,k2 +1 − ci=2 ei . We have α ≥ ur,p(k2 ),k2 + δr (g1 ) + c and hence α ≥
δr (g1 ).
Lemma 3 Let Y ⊂ Pr be a general union of a smooth curve of genus g1 and
degree α, c−1 smooth rational curves of degree e2 , . . . , ec and vr,p(k2 +1)+c−g1 ,k2 +1
lines. Then hi (IZ∪Y (k2 + 1)) = 0, i = 0, 1.
21
Adding a General Union of a Prescribed Number...
Proof. Since k2 > max{|η|, 1}, we have |ur,p(x)+1,k − ur,p(x)+η,x | ≤ 1 for
kr−2
2
x = k2 and x = k2 + 1 and ur,p(k2 +1)+1,k2 +1 − ur,p(k2 )+1,k2 ∼ r·(r−2)!
, we get
α ≥ ur,p(k2 )+1,k2 + δr (g1 ) + 4k. Fix Y = Y1 t E as in Lemma 2 with Y1 a
general element of Z(r, ur,p(k2 )+1,k2 − vr,p(k2 )+1,k2 , g1 ) and E a disjoint union of
vr,p(k2 )+1,k2 lines. We have hi (IZ∪Y (k2 )) = 0, i = 0, 1 (Lemma 2). Without
losing generality we may also assume that Y is transversal to H and that Y ∩H
is formed by deg(Y ) general points of H (Lemmma 1).
(a) First assume vr,p(k2 +1)+η,k2 +1 ≥ vr,p(k2 )+1,k2 . By the inductive assumption with only genera zero if r > 4 or [1] if r = 4, there are is curve W ⊂ H
with the following properties: h1 (H, I(Z∩H)∪W,H (k2 + 1)) = 0, W = W1 t · · · t
Wc+1 , W ∩ Z = ∅, Wc+1 is a disjoint union of vr,p(k2 +1)+η,k2 +1 − vr,p(k2 )+1,k2
lines, Wc+1 ∩ (Y ∩ H) = ∅, Wj , 2 ≤ j ≤ c, is a smooth rational curve of
degree ej with Wj ∩ (Y ∩ H) = ∅, P
W1 is a smooth rational curve of degree ur,p(k2 +1)+c−g1 ,k2 +1 − ur,p(k2 )+1,k2 − ci=2 ei − (vr,p(k2 +1)+η,k2 +1 − vr,p(k2 )+1,k2 ),
W1 ∩E = ∅, and ](W1 ∩Y1 ) = g1 +1. We need to check that we may achieve the
last condition. It is sufficient to use that Y1 ∩ H contains at least g1 + 1 points
of H and that a general rational curve C ⊂ H with deg(C) = deg(W1 ) and C
passes through g1 +1 general points of H. The first condition is satisfied by the
case g = 0 of Remark 1 (we even have that Y1 ∩ H is general in H and hence
it is sufficient to notice that deg(Y1 ) ≥ g1 + 1 by our assumption on k2 . The
second condition is satisfied by the case g = 0, n = r − 1 of Remark 1, because
our assumption on k2 implies deg(W1 ) ≥ g1 + 1. Set F := Y ∪ W . The curve
F is a disjoint union of a nodal curve with arithmetic genus g1 , c − 1 smooth
rational curves of degree e2 , . . . , ec and vr,p(k2 +1)+c−g1 lines. From (3) we get
2
2
h0 (O((Z∪Y )∩H)∪W (k2 + 1)) = r+k
. Hence h0 (O(Z∩H)∪W (k2 + 1)) ≤ r+k
. If
r−1
r−1
1
r = 4, then Z ∩H = ∅; since κ ≥ 5, by [1] we have h (H, I(Z∩H)∪W,H (k2 +1)) =
0. If r > 4, then h1 (H, I(Z∩H)∪W,H (k2 + 1)) = 0, because k2 + 1 ≥ κ
Since Y ∩ H \ W has cardinality ur,p(k2 )+1,k2 − g1 − 1 and it is general in
H, h1 (H, I(Z∩H)∪W,H (k2 + 1)) = 0 and Z ∩ H has p(t) − p(t − 1) as its Hilbert
polynomial, (3) gives hi (H, I((Z∪Y )∩H)∪W (k2 + 1)) = 0, i = 0, 1. Castelnuovo’s
inequalities give hi (IZ∪Y ∪W (k2 + 1)) = 0, i = 0, 1. The semicontinuity theorem
for cohomology gives the lemma when vr,p(k2 +1)+η,k2 +1 ≥ vr,p(k2 )+1,k2 .
(b) Now assume vr,p(k2 +1)+c−g1 k2 +1 < vr,p(k2 )+1,k2 . Write E = E1 tE2 with
E1 a disjoint union of vr,p(k2 +1)+c−g1 ,k2 +1 lines. By the inductive assumption
with only genera zero if r > 4 or [1] if r = 4, there are a curve W ⊂ H with
h1 (H, I(Z∩H)∪W,H (k2 + 1)) = 0, W = W1 t · · · t Wc with Wj , 2 ≤ j ≤ c, a
smooth rational curve of degree ej , W ∩ (Z ∩ H) = ∅, Wj ∩ Y = ∅ for all
j 6= 1, W1 ∩ E1 = ∅, ](W1 ∩ Y1 ) = g1 + 1 and E2 ∩ H ⊂ W1 . We need to check
that W1 contains β := 1 + g1 + vr,p(k2 )+1,k2 − vr,p(k2 )+1,k2 general points of H.
This is true, because β ≤ 2k2 and deg(W1 ) ≥ 2k2 . We use Y ∪ W and the
22
Edoardo Ballico
semicontinuity theorem.
Lemma 4 Fix any integer k ≥ k2 + 2 and any integers b1 , . . . , bc such that
bi ≥ δr (gi ) for all i 6= 1, b1 ≥ α (where α is defined in Lemma 3), b1 +· · ·+bc =
ur,p(k)+η,k − vr,p(k)+η,k . Let Y = Y1 t · · · t Yc t Yc+1 ⊂ Pr be a general union
with Yi ∈ Z(r, bi , gi ), 1 ≤ i ≤ c, and Yc+1 a union of vr,p(k)+η disjoint lines.
Then hi (IZ∪Y (k)) = 0, i = 0, 1.
Proof. Fix k and the integers bi ≥ δr (gi ) for all i 6= 1, b1 ≥ α ,
b1 + · · · + bc = ur,p(k)+η,k − vr,p(k)+η .
(a) Assume k = k2 + 2. Take the set-up of Lemma 3. Recall that ei = 1
if gi = 0 and ei = δr (gi ) − 1 if gi > 0, 2 ≤ i ≤ c. Hence bi − gi ≥ ei ≥ r if
0
gi > 0. Take Y 0 = Y10 t · · · t Yc+1
with Y 0 ∩ Z = ∅, hi (IZ∪Y 0 (k2 + 1)) = 0,
i = 0, 1, Yi0 ∩ Yj0 = ∅ for all i 6= j, Y10 ∈ Z(3, α, g1 ), Yi0 ∈ Z(r, ei , 0) for all
0
i = 2, . . . , c and Yc+1
a disjoint union of vr,p(k2 +2)+η,k2 +2 lines. Since α ≥ 2rg1 ,
we may assume that Y 0 ∩ H is a general subset of H with cardinality deg(Y 0 ).
(a1) First assume vr,p(k2 +2)+η,k2 +2 ≥ vr,p(k2 +1)+c−g1 ,k2 +1 . Take the curve
W = W1 t· · ·tWc+1 ⊂ H in the following way. Wi ∩Wj = ∅ for all i 6= j; Wc+1
is a disjoint union of vr,p(k2 +2)+η,k2 +2 − vr,p(k2 +1)+c−g1 ,k2 +1 lines; Wi , 2 ≤ i ≤ c,
is a smooth rational curve of degree bi − ei containing containing exactly one
point of Yi0 if bi > ei , while Wi = ∅ if bi = ei (in the latter case we have gi = 1
and bi = 1); W1 is a smooth rational curve contained exactly one point of Y10 .
2 +1
From (3) we get h0 (O((Z∪Y 0 )∩H∪W (k2 + 2)) = r+k
. Since Y 0 ∩ H is general
r−1
in H, it is sufficient to prove that h1 (H, I(Z∩H)∪W (k2 + 2)) = 0. If r = 4, then
Z ∩ H = ∅ and hence we only need to prove that h1 (H, IW,H (k)) = 0 with
W ⊂ H a general union of at most c general rational curves with prescribed
degrees. If r = 4, then Z ∩ H = ∅; we use [1] and that κ ≥ 5. If r > 4, then
we use the definition of κ, i.e. inductive assumption in H ∼
= Pr−1 for H ∩ Z,
c, and gi = 0 for all i. Hence from (3) and the generality of the set Y 0 ∩ H we
get hi (H, I((Z∪Y )∩H)∪W,H (k2 + 2)) = 0, i = 0, 1. The Castelnuovo’s inequalities
give hi (IZ∪Y 0 ∪W (k2 + 2)) = 0, i = 0, 1. Use the semicontinuity theorem for
cohomology ([4, Theorem III.12.8]).
0
(a2) Now assume vr,p(k2 +2)+η,k2 +2 < vr,p(k2 +1)+c−g1 ,k2 +1 . Write Yc+1
=
E1 t E2 with deg(E1 ) = vr,p(k2 +1)+c−g1 ,k2 +1 − vr,p(k2 +2)+η,k2 +2 . We make the
construction of step (a1) taking Wc+1 = ∅, Wi as in step (a1) for i = 1, . . . , c
as in step (a1), except that E1 ∩ H ⊂ W . This is possible for the following
reason. Since each component of E1 is a line and E1 ⊂ Pr , E1 ∩ H are deg(E1 )
general points of H; we have deg(E1 ) ≤ k2 , W is a disjoint union of “ general ” rational curve and deg(W ) ≥ rk2 by (3) and our assumption on k2 . Use
this new Y 0 ∪ W and the semicontinuity theorem for cohomology ([4, Theorem
Adding a General Union of a Prescribed Number...
23
III.12.8]).
(b) Now assume k > k2 + 3 and that the lemma is true for the integer
k − 1. Fix bi ≥ δr (gi ) for all i 6= 1, b1 ≥ α , b1 + · · · + bc = ur,p(k)+η,k − vr,p(k)+η .
By the definition of α we also have b1 ≥ δr (g1 ).
Claim 1: There are integers a1 ≥ α and ai ≥ δr (gi ), i = 2, . . . , c, such
that a1 + · · · + ac = ur,p(k−1)+η,k−1 − vr,p(k−1)+η,k−1 and ai ≤ bi for all i.
Proof of Claim 1: We have c − g1 + k2 ur,p(k2 )+c−g1 ,k2 + vr,p(k2 )+c−g1 ,k2 =
with 0 ≤ vr,p(k2 )+c−gP
1 ,k2 ≤ k2 − 1 and
Pc ur,p(k2 )+c−g1 ,k2 − vr,p(k2 )+c−g1 ,k2 =
c
α + e2 + · · · + ec = α + i=2 δr (gi ) − ( i=2 gi . Since k − 1 > k2 , we get
(k − 1)ur,p(k2 )+c−g1 ,k2 − vr,p(k2 )+c−g1 ,k2 ≤ r+k−1
− 2k − 2(k − 1)(g2 + · · · + gc ).
r
Hence (k −1)(ur,p(k2 )+c−g1 ,k2 −vr,p(k2 )+c−g1 ,k2 +g2 +· · ·+gc . We start with the cple (α, δr (g2 ), . . . , δr (gc )) and then increase in each step by one one of its entries,
with the only restriction that after this step for all i = 1, . . . , c the i-th component is at most bi . We make β steps, where β := ur,p(k−1)+η,k−1 − vr,p(k−1)+η,k−1 .
After β steps we get an c-ple (a1 , . . . , ac ) with the properties of Claim 1.
t+k2
r
00
By the inductive assumption there is a disjoint union Y 00 = Y100 t · · · t Yc+1
00
with Yi ∈ Z(r, ai , gi ) if i ≤ c, Yc+1 a disjoint union of vr,p(k−1)+η,k−1 lines,
Y 00 ∩ Z = ∅ and hi (IZ∪Y 00 (k − 1)) = 0, i = 0, 1.
(b1) First assume vr,p(k)+η,k ≥ vr,p(k−1)+η,k−1 . We add W1 t · · · t Wc+1 ⊂
H, with Wc+1 a union of vr,p(k)+η,k vr,p(k−1)+η,k−1 disjoint lines, each Wi , 1 ≤
i ≤ c, a smooth rational curve containing exactly one point of Yi00 ∩ H (case
bi > ai ) or Wi = ∅ if bi = ai . As in step (a1) we get hi (IZ∪Y 00 ∪W (k)) = 0,
i = 0, 1.
0
(b2) Now assume vr,p(k)+η,k < vr,p(k−1)+η,k−1 . Write Yc+1
= E1 t E2 with
deg(E1 ) = vr,p(k)+c−g1 ,k − vr,p(k−1)+η,k−1 Now we add W = W1 t · · · t Wc ⊂ H,
with each Wi a smooth rational curve containing exactly one point of Yi00 ∩ H
(case bi > ai ) or Wi = ∅ if bi = ai ; we also impose that each point of E1 ∩ H is
contained in W . As in step (a2) this is possible, because deg(W ) ≥ r deg(E1 )
(Remark 1).
Proof of Theorem 2: Fix an integer k3 such that k3 ≥ k2 and ur,p(k3 −1)+η,k3 −1 .
We claim that we may take k0 := k3 + 5; we may also take k0 = k2 + 5 if we
only look at integers b1 , . . . , bc with the additional condition that b1 ≥ α. Fix
an integer k ≥ k2 + 5. Fix integers bi ≥ δr (gi ), 1 ≤ i ≤ c, b1 ≥ α, such that the
union of Z and the disjoint union of c elements of Z(r, bi , gi ), i = 1, . . . , c, has
critical value k. Since Z(r, d1 , g1 )×· · ·×Z(r, bc , gc ) is irreducible, it is sufficient
24
Edoardo Ballico
to find A = A1 t · · · t Ac and B = B1 t · · · t Bc with A ∩ Z = B ∩ Z = ∅,
Ai ∈ Z(r, bi , gi ) for all i, Bi ∈ Z(r, bi , gi ) for all i, h1 (IZ∪A (k)) = 0 and
h0 (IZ∪B (k − 1)) = 0. Set b := b1 + · · · + bc . By the definition of critical
value and the inequalities k − 1 ≥ κ we have h0 (OZ (k)) + kd + η ≤ r+k
,
r
r+k−1
0
h (OZ (k − 1) + (k − 1)d + η >
and ur,p(k−1)+η,k−1 < d ≤ ur,p(k)+η,k .
r
(a) In this step we prove the existence of the curve A.
(a1) First assume d ≥ ur,p(k−1)+η,k−1 + rvr,p(k−1)+η,k−1 . As in the proof of
Claim 1 of the proof of Lemma 4 there are integers a1 , . . . , ac such that δr (gi ) ≤
ai ≤ bi for all i, a1 ≥ α and a1 + · · · + ac = ur,p(k−1)+η,k−1 − vr,p(k−1)+η,k−1 . By
Lemma 4 applied to the integers k −1 and a1 , . . . , ac there is Y = Y1 t· · ·tYc+1
with Y ∩ Z = ∅, Yi ∈ Z(r, ai , gi ) for all i ≤ c, Yc+1 a disjoint union of vr,p(k−1)+η
lines and hi (IZ∪Y (k − 1)) = 0, i = 0, 1. Let W = W1 t · · · t Wc a general
union of smooth rational curves of degree bi − ai (the ∅ if bi = ai ) with the only
restriction that if bi > ai (i = 1, . . . c), then Wi contains exactly one point of
Yi and Yc+1 ∩ P
H ⊂ W ; we use Remark 1 or Lemma 1 to satisfy the last condition, because i (bi − ai ) ≥ rvr,p(k−1)+η,k−1 . The curve Y ∪ W has c connected
components, Y1 ∪ Wc , . . . Yc ∪ Wc , and Yi ∪ Wi ∈ Z 0 (r, bi , gi ) for all i. Hence
by the semicontinuity theorem for cohomology it is sufficient to prove that
h1 (IZ∪Y ∪W (k)) = 0. Since ResH (Z ∪ Y ∪ W ) = Z ∪ Y and hi (IZ∪Y (k − 1)) = 0,
i = 0, 1, it is sufficient to prove that h1 (H, I((Z∪Y )∩H)∪W (k)) = 0. Since
d ≤ ur,q(k)+η,k , (3) gives h0 (O((Z∪Y )∩H)∪W (k)) ≤ r+k−1
− vr,p(k)+η,k and hence
r−1
r+k−1
0
h (O((Z∪Y )∩H)∩W (k)) ≤ r−1 . Since H ∩Y is a general subset of H with cardinality deg(Y ) (+++), it is sufficient to prove that h1 (H, I(Z∩H)∪W (k)) = 0. If
r = 4, then Z∩H = ∅ and hence we only need to prove that h1 (H, IW,H (k)) = 0
with W ⊂ H a general union of at most c general rational curves with prescribed degrees. Since we use the definition of κ, i.e. we use the inductive
assumption in HPr−1 for H ∩ Z, c, and gi = 0 for all i.
(a2) Assume d < ur,p(k−1)+η,k−1 + rvr,p(k−1)+η,k−1 . By (3) and the inequal
ity ur,p(k)+η,k − ur,p(k−1)+η,k−1 ≥ 2rk we have h0 (OZ (k) + kd + η ≤ r+k
−k
r
r+k
0
and hence h (OZ (k) + kd + η ≤ r − vr,p(k−1)+η,k−1 . Since d ≥ ur,p(k−1)+η,k−1
there are integers u1 , . . . , uc such that bi ≥ ui ≥ ci for all i, u1 ≥ α, ui ≥ δr (gi )
for all i and u1 + · · · + uc = ur,p(k−1)+η,k−1 . We modify step (a) of the proof of
Lemma 4 (here from k − 2 to k − 1) in the following way. Since k − 2 ≥ k2 + 3,
there are integers w1 , . . . , wc such that u1 ≥ w1 ≥ α, and ui ≥ wi ≥ δr (gi )
0
0
for all i ≥ 2. Take a solution Y 0 = Y10 t · · · t Yc+1
with Yc+1
a disjoint union
0
0
0
of vr,p(k−2)+η,k−2 . Let W = W1 t · · · t Wc ⊂ H be a general of c rational
curves of degree ui − wi (or the ∅ if ui = wi ), with Wi0 containing exactly
one point of Yi00 if Wi 6= ∅, so that Y 00 ∪ W 0 has c connected components,
Y100 ∪ W10 , . . . , Yc00 ∪ Wc0 with each Yi00 ∪ Wi0 ∈ Z 0 (r, ui , gi ). Using [1] (if r = 4)
or the inductive assumption (if r > 4) we get h1 (IZ∪Y 00 ∪W 0 (k − 1)) = 0. Let
Adding a General Union of a Prescribed Number...
25
T = T1 t · · · t Tc a general smoothing of Y 00 ∪ W 0 . By the semicontinuity
theorem we have h1 (IZ∪T (k − 1)) = 0. Let F = F1 t · · · t Fc ⊂ H be a general
union of smooth rational curves of degree b1 − u1 , . . . , bc − uc (the ∅ if bi = ui )
with the only restriction that if bi − ai > 0, then Fi contains exactly one point
of Ti . The curve T ∪ F has c connected components, T1 ∪ F1 , . . . , Tc ∪ Fc with
Fi ∈ Z 0 (r, bi , gi ) for all i.
(b) In this step we prove the existence of the curve B. We modify
step (a2) in the following way. Since b1 ≥ α, bi ≥ δr (gi ) for all i and
b1 + · · · + bc > ur,p(k−1)+η,k−1 ≥ ur,p(k−2)+η,k−2 as in the proof of Claim 1
in the proof of Lemma 4 we get the existence of integers zi , 1 ≤ i ≤ c, such
that z1 ≥ α, bi ≥ zi ≥ δr (gi ) for all i and z1 + · · · + zc = ur,p(k−2)+η,k−2 −
00
vr,p(k−2)+η,k−2 . By Lemma 4 there is Y 00 = Y100 t · · · t Yc+1
with Y 00 ∩ Z = ∅,
00
00
Yc+1 a disjoint union of vr,p(k−2)+η,k−2 lines, Yi ∈ Z(r, zi , gi ) for all i ≤ c and
hi (IZ∪Y 00 (k − 2)) = 0, i = 0, 1. Let W 00 = W100 t · · · t Wc00 ⊂ H be a general
of c rational curves of degree bi − zi (or the ∅ if bi = zi ), with Wi00 containing exactly one point of Yi00 if Wi 6= ∅, so that Y 00 ∪ W 00 has c connected
components, Y100 ∪ W100 , . . . , Yc00 ∪ Wc0 with each Yi00 ∪ Wi00 ∈ Z 0 (r, bi , gi ), and
00
∩ H ⊂ W 00 (see step (a1) for the latter condition) . By the Castelnuovo’s
Yc+1
inequalities it is sufficient to prove that h0 (H, I((Z∪Y 00 )∩H)∪W 00 (k − 1)) = 0.
Using [1] (if r = 4) or the inductive assumption (if r > 4) we get that
either h1 (H, I(Z∩H)∪W 00 (k − 1)) = 0 or h0 (H, I(Z∩H)∪W 00 (k − 1)) = 0. If
h0 (I(Z∪H)∪W 0 (k − 1)) = 0, then h0 (H, I((Z∪Y 00 )∩H)∪W 00 (k − 1)) = 0. Hence we
may assume h1 (H, I(Z∩H)∪W 00 (k − 1)) = 0. Hence h0 (H, I(Z∩H)∪W 00 (k − 1)) =
r+k−2
− p(k − 1) + p(k − 2) − h0 (OW 00 (k − 1)). Call β the last integer. By (3)
r−1
for the integer k − 1 we have β = −(k − 1)(d − ur,p(k−1)+η,k−1 ) + vr,p(k−1),k−1 +
](Y 00 ∩ H) − ](Y 00 ∩ W 00 ). Since Y 00 ∩ H \ Y 00 ∩ W 00 is general in H, we get
h0 (H, I((Z∪Y 00 )∩H)∪W 00 (k − 1)) = 0.
(c) Now assume k ≥ k3 + 5. From steps (a) and (b) we conclude if we
assume that b1 , . . . , bc satisfies bi ≥ δr (gi ) and b1 ≥ α. Take arbitrary integers
wi ≥ δr (gi ) such that w1 + · · · + wc ≥ ur,p(k−1)+η,k−1 . By the definition of the
integer k3 there is at list one integer j ∈ {1, . . . , c} such that bj ≥ α. We may
rewrite the proof with the same k1 and k2 (we used bound on g1 + · · · + gc , not
on g1 ).
Proof that Theorem 2 for r, Z, c, gi implies Theorem 1 for the
same data Fix integers r ≥ 4, c > 0, gi ≥ 0 and an admissible polynomial
p(t) (i.e. the Hilbert polynomial of some closed subscheme of some projective
space) such that deg(p(t)) ≤ r − 4. Let Z ⊂ Pr be a closed subscheme with
p(t) as its Hilbert polynomial. By Gotzmann’s theorem there in an integer t0
such that hi (OZ (t + 1 − i)) = 0 for all t ≥ t0 and all i > 0 and t0 depends
26
Edoardo Ballico
only on the polynomial p(t). We take t0 = 0 if Z = ∅. Fix integers bi ≥ δr (gi ),
1 ≤ i ≤ c, and set d := b1 +·· ·+bc . Set η := c−g1 −· · ·+gc . We assume that d
0
is very large, e.g. d > c r+t
+ g1 + · · · + gc . With these assumptions it is easy
r
to check that a general disjoint union Y = Y1 t· · ·tYc , Yi ∈ Z(r, bi , gi ) satisfies
h0 (IY (t0 + 1)) = 0. Hence to study the function t 7→ h0 (IZ∪Y (t)) it is sufficient
to study it when t > t0 . We say that (r, p(t), c, g1 , . . . , gc , b1 , . . . , bc ) has critical
value k if k is the minimal integer > t0 such that kd + η + p(k) ≤ r+k
,
r
i.e. d ≤ ur,p(k)+η,k . Since ur,p(k)+η,k > ur,p(k−1),k−1 (use (3)), we get that
(r, p(t), c, g1 , . . . , gc , b1 , . . . , bc ) has critical value k if and only if ur,p(k−1),k−1 <
d ≤ ur,p(k)+η,k . Assume that (r, p(t), c, g1 , . . . , gc , b1 , . . . , bc ) has critical value
k. Fix Y = Y1 t · · · t Yc , Yi ∈ Z(r, bi , gi ), with Z ∩ Y = ∅. If Z ∪ Y has
maximal rank, then h0 (IZ∪Y (k − 1)) = 0 and h1 (IZ (k)) = 0. Now assume
that h0 (IZ∪Y (k − 1)) = 0 and h1 (IZ (k)) = 0. Obviously h0 (IZ∪Y (t)) = 0 for
all t ≤ k − 1. Our assumptions on the integers bi imply h1 (OY (1)) = 0. Since
k ≥ max{2, t0 }, Castelnuovo-Mumford’s lemma implies h1 (IZ∪Y (t)) = 0 for
all t > k. Take k3 as in the proof of Theorem 2 and take ∆ := ur,p(k3 +1),k3 +1 .
Take bi ≥ δr (gi ), i = 1, . . . , c, such that d := b1 + · · · + bc ≥ ∆. Obviously
(r, p(t), c, g1 , . . . , gc , b1 , . . . , bc ) has critical value k > k0 . Fix a general Y =
Y1 t · · · t Yc be a general union with Yi ∈ Z(r, bi , gi ). In particular Z ∩ Y = ∅.
Theorem 2 for the data p(t), c, gi gives that h1 (IZ∪Y (k)) = 0. Since k − 1 ≥ k3 ,
it gives h0 (IZ∪Y (k − 1)) = 0. Hence Z ∪ Y has maximal maximal rank by the
Castelnuovo-Mumford’s lemma).
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27
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